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  • Two point charges -q and +q are placed at a distance of L, as shown in the figure.  <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/01/31/capturefgr5.PNG" style="heig

Two point charges -q and +q are placed at a distance of L, as shown in the figure.

 

The magnitude of electric field intensity at a distance \mathrm{R(R> > L)} varies as

Option: 1

\frac{1}{R^{3}}


Option: 2

\frac{1}{R^{4}}


Option: 3

\frac{1}{R^{6}}


Option: 4

\frac{1}{R^{2}}


Answers (1)

best_answer

\mathrm{E_{1}=\frac{k q}{R^{2}}} \\

\mathrm{E_{2}=\frac{k q}{(R+L)^{2}}} \\

\mathrm{E_{\text {net }}=E_{1}-E_{2}}

         \mathrm{=\frac{k q}{R^{2}}-\frac{k q}{R^{2}\left(1+\frac{L}{R}\right)^{2}}} \\

        \mathrm{=\frac{k q}{R^{2}}\left(1-\left(1+\frac{L}{R}\right)^{-2}\right]}

\mathrm{E_{n e t}=\frac{2 k q L}{R^{3}}}

         \mathrm{=\frac{k q}{R^{2}}\left(1-\left(1-\frac{2 L}{R}\right)\right)} \\            \mathrm{(\text { by binomial expantion }})

Hence correct option is 1

Posted by

manish

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